Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{2(3k - 2)}{k} \div \dfrac{5(3k - 2)}{3k} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{2(3k - 2)}{k} \times \dfrac{3k}{5(3k - 2)} $ When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 2(3k - 2) \times 3k } { k \times 5(3k - 2) } $ $ y = \dfrac{6k(3k - 2)}{5k(3k - 2)} $ We can cancel the $3k - 2$ so long as $3k - 2 \neq 0$ Therefore $k \neq \dfrac{2}{3}$ $y = \dfrac{6k \cancel{(3k - 2})}{5k \cancel{(3k - 2)}} = \dfrac{6k}{5k} = \dfrac{6}{5} $